Indefinite integration class 12

PROBABILITY
Introduction (Terminology):
In physics and chemistry, we perform experiments and in
each case result is same when repeated under identical
conditions.
But, in statistical experiments the result may be
altogether different, even when they are performed
under identical conditions.
Random experiment or Trial:

PROBABILITY
Ex: (1) Tossing a coin is a random experiment as the
result is either head(H) or tail(T) but not known
exactly.
(2) Throwing a die is a random experiment as the
result is 1 or 2 or………….. or 6, but not known exactly.
An experiment is said to be random experiment, if the
result is not certain but it may be one of the several
possible outcomes, when repeated under identical
conditions.

PROBABILITY
Sample space(S):
The set of all distinct possible outcomes of a random
experiment is called the sample space (denoted by S) of
random experiment.
Also n(S) denotes number of elements in sample space S.
Each outcome of S is called a sample point.
Example :
(1) In case of tossing a coin , S={H,T}
(i.e.) n(S)=2

PROBABILITY
(2) In case of tossing 2 coins , S={HH,HT, TH, TT}
(i.e.) n(S)= 2𝟐
=4
(3) In case of tossing 3 coins ,
S={HHH,HHT, HTH, THH,TTH,THT,HTT,TTT}
3 heads 2 heads 1 head 0
heads
Also n(S)= 2𝟑
=8
(4) If K coins are tossed at a time,Then n(S)= 2𝒌

PROBABILITY
(5) Throwing a single die gives S={1,2, 3,4,5,6} (i.e) n(S)=6
(6) Throwing two dies at a time gives
S=
(1,1), (1,2), (1,3),……. (1,6)
(2,1), (2,2), (2,3),……. (2,6)
⋮ ⋮ ⋮ ⋮
(6,1), (6,2), (6,3),……. (6,6)
Thus n(S)= 6𝟐
=36
(7) If k dies are rolled at a time, then n(S)= 6𝒌

PROBABILITY
(8) An urn contains 3 white, 4 red and 7 green balls
(i.e) 14 balls in total. If 5 balls are drawn at random from
urn, then n(S) = 14C𝟓
MCQ
When a coin and a die are rolled together, Then n(S)=
a) 6𝟐 b) 2𝟔
c) 12 d) 2𝟏𝟐

PROBABILITY
Event:
Any subset of the sample S is called an event.
Example:
(1) By throwing a single die, sample space= S={1,2,3,4,5,6}.
Consider A ={2,4,6}, B={1,3,5}. Now A, B are subsets of S
and hence they are events.
In case the subset is a singleton set, then we call it
as simple event. Usually events are denoted by
capital letters A,B,C…..

PROBABILITY
(2) 4 cards drawn from a well shuffled pack of 52 cards. Let
A be the event that all 4 cards red, then n(A) = 26C𝟒
Assume that B denotes that an event that all 4 cards are
of same face value,
A is the event of getting an even number and B is
the event of getting an odd number.
we know that there are 13 sets ( each set has 4 cards)
having same face value. Now select one set from 13 sets
in 13C𝟏
ways.
(i.e) n(B)=13C𝟏

PROBABILITY
Certain event & Impossible event:
An event which is certain to happen is called a certain event
and an event which can’t happen at all is called an impossible
event.
Example:
When a usual die is rolled
(1) Getting a multiple of 7 is an impossible event.
(2) Getting a number less than 7 is a certain event.

PROBABILITY
Complement of an Event:
S is sample space of random experiment and A be any event
(i.e) A ⊆ S, Then Ac = S−A is called
the complement of A , it may be denoted by A
Example:
(1) If a die is rolled, Then S={1,2,3,4,5,6}
A=event of getting multiple of 3={3,6}
Now A=S-A={1,2,4,5}.

PROBABILITY
(2) If a card is drawn from a pack of 52 cards and A is the event of
getting king card, Then A is the event of not getting king
card. Hence A contains 4 sample points, where as A contains
48 sample points.
Note:
(De Morgan’s laws)
(1) A ∪ B = A ∩ B
(2) A ∩ B = A ∪ B

PROBABILITY
Compound event (Simultaneous occurrence of the events):
A,B are 2 events associated with a random experiment, then (A ∩
B) is called as compound event
A card is selected at random from a
well shuffled pack of 52 cards.
Example:
A=Event of getting the card as red card
B=Event of getting the card as king
Now A ∩ B = card is red and king
In red cards there are two kings 1) hearts king 2) diamond king
and (A ∩ B) occurs if and only
if both A and B have occurred simultaneously.

PROBABILITY
We have n(A)= 26C𝟏 =26; n(B)=
4C𝟏 =4; n(A ∩B)= 2C𝟏 =2
Mutually exclusive events:
Events are said to be mutually exclusive, if the happening of
any one of them prevents the happening of any of the other
events.
If A, B are 2 events in sample space; then they are said to be
mutually exclusive iff A∩B= 𝛟
Example:
If a die is rolled, then S={1,2,3,4,5,6}.

PROBABILITY
A= events of getting an even number={2,4,6}
B= events of getting an odd number={1,3,5}
Observe that A∩B= 𝛟 A,B are mutually exclusive.
Note:
Events are such that happening of one of them assures
others do not happen. Then, those events are always
mutually exclusive.
Mutually exclusive events are nothing but disjoint events.

PROBABILITY
Two or more events said to be equally likely
or equiprobable, if there is no reason to expect
any one of them to happen in preference to the others.
Example:
By throwing a usual die, there is an equal chance to get
any of the 6 faces to turn up. Therefore, we say that the
six simple events are equally likely to happen.
Equally likely events:

PROBABILITY
E1∪ E2 ∪ ……. ∪ Ek =S (sample space)
Example:
Two coins are tossed, then S={HH,HT,TH,TT}.
Exhaustive Events:
Two or more events are said to be exhaustive, if the
performance of the experiment always results in the
occurrence of at least one of them. Thus events E1, E2,…….
Ek are said to be exhaustive if
Let A,B,C are events of getting 2 heads, 1 head , 0 heads
respectively
(i.e) A={HH}, B={HT,TH}, C={TT}. Observe that A∪ B ∪ C=S
⟹ A, B, C are exhaustive events.

PROBABILITY
Classical or mathematical definition of probability:
A random experiment results n simple events which are
exhaustive, mutually exclusive and equally likely. In those n
events , m are favourable for the occurrence of the event E;
then the ratio
m
n
is defined as the probability of the event E,
denoted by P(E). Thus
P(E)=
m
n =
no.of favourable outcomes to E
no.of all possible out comes

PROBABILITY
n(E) stands for number of sample points favouring event E
n(S) stands for number of sample points in sample space S.
Remarks:
(1) From the definition it is clear that,
0≤ P(E) ≤ 1 for any event E
(2) If P(E)=0, then E is an impossible event
(3) If P(E)=1, then E is a sure event.
(4) By definition P(E)=
m
n
We write P(E)=
n(E)
n(S)
, where

PROBABILITY
⇒ m are favouring E and (n−m) are not in favour of E (denoted by
E)
∴ P(E)=
n−m
n
=1−
m
n
=1−P(E)
⇒ P(E)=1−P(E) ⇒ P(E)+P(E)=1
(5) The ratio m: (n-m) or P(E):P(E) is called as odds infavour of
event E. Also the ratio (n-m): m (or) P(E):P(E) is called as
odds against event E.

PROBABILITY
Limitations of classical definition:
Above classical definition is not applicable when
(1) Outcomes of random experiment are not equally similar
(2) The random experiment contains infinitely many
outcomes in order to overcome these difficulties,
we define the probability in another way based on
relative frequencies as follows:

PROBABILITY
Suppose E is an event of a random experiment
which can be repeated any number of times under similar
conditions; noting the success or failure of the event in
each trial. If Nr(E) denotes the number of successes in first
r trials, then
Statistical definition of probability:
we call
Nr(E)
r
, The relative frequency and denote it by R𝒓(E)
and get the relative frequencies R𝟏(E) ,R𝟐(E), … … . . R𝒓(E). If
Lt
n→∞
Rn(E) exists, this limit is defined as probability of E,
denoted by P(E).

PROBABILITY
So far practical purpose, we take n as sufficiently large and
R𝒏(E) is taken for P(E).
Since every sequence need not be convergent, there is no
guarantee that Lt
𝒏→∞
R𝒏(E) exists.

PROBABILITY
Example:
Suppose a die is thrown1000 times and is observed that
1,2,3,4,5 and 6 appears 130,140, 136,134,225 and 235
times respectively. Then we take
Pr(1)=
130
1000
= 0.13,
Pr(2)=
140
1000
= 0.14
Pr(3)=
136
1000
= 0.136,

PROBABILITY
Pr(4)=
134
1000
= 0.134
Pr(5)=
225
1000
= 0.225,
Pr(6)=
235
1000
= 0.235
Observe that P(1)+P(2)+……+P(6)=1

PROBABILITY
Drawbacks of statistical definition of Probability:
i) Repeating random experiment infinitely many times is
practically impossible
ii) The sequence R𝟏(E) ,R𝟐(E), … … . . R𝒏(E) is assumed to tend
a limit, which may not exist.
Keeping the above troubles in mind, Axiomatic
definition is formulated by Russian mathematician
KOLMOGOROV as follows:

PROBABILITY
Axiomatic definition of Probability:
Let S be the sample space (with finite elements) of a
random experiment. P (S) denotes power set of S i.e all
possible subsets S i.e all events. We define probability
function ‘P’ so that P: P (S)R and
(1) P(E)≥0 ∀E𝛜P (S)
(2) P(S)=1

PROBABILITY
(3) E1, E2 𝛜P (S) , E1∩ E2=∅, then P(E1∪E2 )=P(E1)+P(E2)
Any function P defined on P (S) and satisfying above
axioms is called as probability function.
(1) There can be any number of probability functions
for the same random experiment.
Remarks:
(2) Axiom of additivity can be generalized as follows: If
E1, E2, E3,…… En,….are mutually exclusive events (i.e)
Ei∩Ej=∅ (i ≠j), Then

PROBABILITY
P(E1∪E2….…. ∪En ∪….)= P(E1)+ P(E2)+……+ P(En)+…..
(3) Let us prove P(E)+P(E)=1 by axiomatic approach:
We have E∪ E= SP(E∪ E)= P(S)
P(E)+P(E)=1 ∵ E ∩ E = ∅

PROBABILITY
1) In the experiment of throwing a die, consider the
following events:
A={1,3,5}, B={2,4,6}, C={1,2,3} are these events
equally likely?
Solution:
Yes, clearly A,B,C are equally likely due to the fact
P(A)= P(B)= P(C)=
3
6
.

PROBABILITY
2) If two numbers are selected randomly from 20
consecutive natural numbers, find the probability that
the sum of the two numbers is
i) An even number ii) An odd number
Solution:
i) Let A be the event that the sum of the two numbers
is even when two numbers are selected at random
from 20 consecutive natural numbers and S be the
sample space.
Now n(S)= 20C𝟐
= 190

PROBABILITY
Since the sum of two numbers is even if the two numbers
are both even or both odd.
10C𝟐
+ 10C𝟐
= 45+45=90.
n(A)= Now P(A)=
n(A)
n(S)
=
90
190
=
9
19
.
ii) The sum of two numbers is odd (Let , denoted by B) if
one number is even and the other number is odd.
∴n(B)=10C𝟏
× 10C𝟏
= 100
∴P(B)=
n(B)
n(S)
=
100
190
=
10
19

PROBABILITY
Solution:
3) A game consists of tossing a coin 3 times and noting its
outcome. A boy wins if all tosses give the same outcome
and loses otherwise. Find the probability that the boy
loses the game.
Let A be the event that the boy tosses the same
outcome in three tosses(HHH or TTT) and S be the
sample space. Now n(S)=2𝟑
= 8, n(A)=2.
Now P(A)=
n(A)
n(S)
=
2
8
=
1
4
.
The probability that the boy loses the game
=P(A)=1-P(A)=1−
1
4
=
3
4
.

PROBABILITY
Solution:
4) On a festival day, a man plans to visit 4 holy temples
A,B,C,D in a random order. Find the probability that he
visits i) A before B ii) A before B and B before C.
i) Let E be the event that the man visits A before
B and S be the sample space.
Now n(S)=4! = n(E) =
4!
2!
.
∴ P(E) =
n(E)
n(S)
=
4!/2!
4!
=
1
2
.

PROBABILITY
ii) Let E be the event that the man visits A before B, B
before C and S be the sample space.
Now n(S)=4! = n(E) =
4!
3!
.
∴ P(E) =
n(E)
n(S)
=
4!/3!
4!
=
1
6
.

PROBABILITY
∴ P(A) =
n(A)
n(S)
=
2550
4950
=
17
33
.
ii) Let A be the event that both enter the different
sections and S be the sample space.
Now n(S)= 100C𝟐
=
100×99
2
= 4950,
n(A)= 40C𝟏
× 60C𝟏
= 2400.
∴ P(A) =
n(A)
n(S)
=
2400
4950
=
16
33
.

PROBABILITY
Solution:
5) Out of 100 students, two sections of 40 and 60 are
formed. If you and your friend are among the 100
students, find the probability that i) you both enter the
same section ii) you both enter the different sections.
i) Let A be the event that both enter the same section
and S be the sample space.
Now n(S)= 100C𝟐
=
100×99
2
= 4950,
n(A)= 40C𝟐
+ 60C𝟐
=
40×39
2
+
60×59
2
= 780+1770=2550

PROBABILITY
MCQ:
1) Identify the wrong statement in the following:
(a) 0 ≤ P(A) ≤ 1
(b) P(E) + P(E) = 1
(c) P(A ∪ B) = P(A)+ P(B)
(d) The ratio P(E): P(E) gives odds in favour of event E

PROBABILITY
(2) When 2 coins are tossed at a time, then probability of
getting 2 heads is__
(a)
1
2
(b)
1
4
(c)
3
4
(d) 0

PROBABILITY
(3) When 2 dice are rolled together, then probability
of getting same number on both the dice is _____
(a)
3
4
(b)
5
6
(c)
3
6
(d)
1
6

PROBABILITY
(4) Two cards are drawn from a well shuffled pack of 52
cards. The probability that one of them is black and the
other is red is__
(a)
13
51
(b)
26
51
(c)
13
102
(d)
13
204
Hint:
n(S)= 52C𝟐
n(E)= 26C𝟏
× 26C𝟏
P(E)=
n(E)
n(S)

Indefinite integration class 12

More Related Content

Similar to Indefinite integration class 12 (20)

More from nysa tutorial (20)

Recently uploaded (20)

Indefinite integration class 12